It means $x$ is a fraction (floating point numbers).
Now $\pi>3\Leftrightarrow \pi^2>9$
Again $\pi<4\Leftrightarrow 2\pi<8\Leftrightarrow2\pi+1<9 $
Combining them yields $2\pi+1<\pi^2 \tag{i}$
Now
$\begin{align}x&< 1 \\ \Rightarrow x^{\pi}&< 1 \tag{ii}\end{align}$
$\begin{align}\therefore x+x^{\pi}&< 2 \\ \Rightarrow (x+x^{\pi})\pi&< 2\pi \end{align} \tag{iii}$
Now from no$\mathrm{(ii)}$ multiplying both sides by $x$, we get
$\begin{align} x\cdot x^\pi &< 1\cdot x \\ \Rightarrow x^{\pi+1} &< x \tag{iv} \end{align}$
no$\mathrm{(iii)}+$no$\mathrm{(iv)}\Rightarrow$
$\begin{align}(x+x^{\pi})\pi+x^{\pi+1} &< 2\pi+ x \\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< 2\pi+x \\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< 2\pi+1; ~[\because x<1 \Rightarrow 2\pi+x< 2\pi+1]\\ \Rightarrow x\pi+(\pi+x)x^{\pi} &< \pi^2; ~[\mathrm{From~no(i)}] \end{align}$
As $x>0$, definitely $\pi^2<x^2+\pi^2+x^{2\pi}$.
$$\therefore x\pi+(\pi+x)x^{\pi} < x^2+\pi^2+x^{2\pi}$$
So, the option C is false. Thus the option A is true.
